# Thread: Cartesian Equation of a Plane

1. ## Cartesian Equation of a Plane

I'm having trouble solving this. I seem to be totally lost. If someone could just tell me the steps I need to take, I'm pretty sure I can work backwards to understand it.

Question: Determine the Cartestian equation of the plane that passes through the points (1, 4, 5) and (3, 2, 1) and is perpendicular to the plane 2x-y+z-1=0.

The answer is 3x+5y-1z-18=0

Thanks

2. Originally Posted by MATHDUDE2
I'm having trouble solving this. I seem to be totally lost. If someone could just tell me the steps I need to take, I'm pretty sure I can work backwards to understand it.

Question: Determine the Cartestian equation of the plane that passes through the points (1, 4, 5) and (3, 2, 1) and is perpendicular to the plane 2x-y+z-1=0.

The answer is 3x+5y-1z-18=0

Thanks
1. The points P(1,4,5) and Q(3,2,1) are determined by the stationary vectors $\displaystyle \vec p$ and $\displaystyle \vec q$.

2. The vector $\displaystyle \vec v = \vec q - \vec p$ and the normal vector of the given plane $\displaystyle \vec n$ span the new plane. Additionally the new plane has to pass through P (or Q).

3. The normal vector of the new plane is $\displaystyle \vec s = \vec v \times \vec n = (-6, -10, 2)$

4. $\displaystyle \vec s \cdot ((x,y,z)-\vec p)=0$ yields the equation of the new plane.

EDIT: I've attached the graphs of the 2 planes. The Points P and Q are drawn in blue and red respectively.

3. thanks!!